What is the Hardy-Weinberg principle?

The Hardy-Weinberg principle, additionally referred to as the Hardy-Weinberg equilibrium, is a collection of 5 presumptions which as soon as satisfied can permit the determination of allele and genotype frequencies that a population. These frequencies will likewise remain continuous for future generations. The principle was uncovered by Godrey Hardy and also Wilhelm Weinberg in 1908, based on Gregor Mendel’s legislation of Segregation. To calculation the frequency that alleles and genotypes that a specific population, over there is two straightforward formula that deserve to be used.

You are watching: In the hardy-weinberg formula, what does 2pq represent?

The presumptions of the Hardy-Weinberg principle

There are 5 presumptions that room made when using the Hardy-Weinberg equations. This are:

No organic selection: There space no evolutionary pressure which may favour a certain allele.Random mating: each individual in a populace mates randomly so the mating v an individual moving a specific allele is not favoured.No mutations: There room no DNA mutations occurring for the alleles i m sorry may affect their function.A closeup of the door population: people within the population do not leave and brand-new individuals room not introduced to the population.Large population size: The populace is considered big enough, at ideal infinite, for this reason that major changes in allele frequencies do not cause a hereditary drift.

If any kind of of these assumptions are not satisfied, then the principle can not be applied.

Determining the allele frequency

The first Hardy-Weinberg equation (p + q = 1) comes to estimating the frequency the alleles in a population. Every gene usually has actually two alleles (diploid organism), one from each parent. These alleles space denoted as the leading (A) and also recessive (a) forms. These are represented as ‘p‘ and also ‘q‘ is the equation below.

In a population, the merged frequency the both the alleles need to equal 1 (100%).

Therefore, if the frequency that one allele is known, it is feasible to calculate the frequency of the various other allele merely by rearranging the equation.


In a population, there are two alleles for ear shape: having actually detached lobes (dominant, A) or having attached lobes (recessive, a). Identify the allele frequency that the recessive allele ‘a’ (attached lobes) given the frequency of the leading allele ‘A’ (attached lobes) is 73%.

1. To figure this out we very first need to fill in what we know into the Hardy-Weinberg equation, i.e. The allele ‘A‘ (p in the equation) frequency is 73% (which is the exact same as 0.73).

2. Next, rearrange the formula to recognize the value of q (the recessive allele frequency). So this would certainly give: q = 1 – 0.73.

3. Calculating this would mean: q = 0.27 (27%). So, 27% that the populace will have actually the allele for attached earlobes.

The recessive allele frequency is 27%.

Determining the genotype frequencies

The Hardy-Weinberg equation provided to recognize genotype frequencies is: p2 + 2pq + q2 = 1.

Where ‘p2‘ to represent the frequency that the homozygous dominant genotype (AA), ‘2pq‘ the frequency of the heterozygous genotype (Aa) and ‘q2‘ the frequency the the homozygous recessive genotype (aa). The amount of these 3 genotypes need to equal 1 (100%).

Again, if one genotype frequency is known, that is possible to use the Hardy-Weinberg equations to work-related out the others.


Let’s usage the same instance above about earlobes (detached lobes and also attached lobes). Recognize the genotype frequency of the homozygous leading (AA) allele for having detached earlobes given the frequency that the attached earlobe (aa) phenotype is 6%.

1. Through looking at the question, we room asked to calculate (AA), for this reason ‘p2‘ in the Hardy-Weinberg equation, given the frequency that aa (‘q2‘ in the equation) is 6% (we will job-related in decimals from this point, therefore this would certainly be 0.06). Due to the fact that ‘q2‘ equates to 0.06, we can work the end what ‘q‘ is by square-rooting 0.06. Act this provides 0.245.

2. Due to the fact that we now understand the allele frequency that the recessive allele (q), we have the right to use the first Hardy-Weinberg equation above (p + q = 1) to work-related out the allele frequency for the leading allele (p).

3. We now understand what p is (0.755). To calculate the genotype frequency that the homozygous leading genotype, we merely need come square the worth of p. Doing this would give 0.57. So, 57% that the population are homozygous leading (AA). This would certainly be the answer come our question.

The homozygous leading genotype frequency is 57%.

4. If you want to go one action further and also work the end the frequency that the heterozygous genotype (Aa), for this reason ‘2pq‘ in the Hardy-Weinberg equation, you have the right to do. Since we recognize the value of ‘p‘ (0.755) and also ‘q‘ (0.245). All that is forced is to multiply 2 by 0.755 and also by 0.245. Doing this will give 0.37. So, 37% that the populace will have actually the heterozygous genotype (Aa).

See more: 1/2 Cup Egg White S 1/2 Cup Egg White, Egg Whites 1/2 Cup Egg White

Notice, if you add all the the genotype frequencies together, this equals 1 (0.57 + 0.37 + 0.06 = 1).